a) CuSO4+2NaOH--->Cu(OH)2+Na2SO4
Cu(OH)2--->CuO+H2O
m\(_{CuSO4}=\frac{200.8}{100}=16\left(g\right)\)
n\(_{CuSO4}=\frac{16}{160}=0,1\left(mol\right)\)
m\(_{NaOH}=\frac{120.10}{100}=12\left(g\right)\)
n\(_{NaOH}=\frac{12}{40}=0,3\left(mol\right)\)
=> NaOH duư
dd sau pư gồm NaOHdư và Na2SO4
m dd sau pư=200+150=350(g)
Theo pthh
n\(_{NaOH}=2n_{CuSO4}=0,2\left(mol\right)\)
n\(_{NaOH}dư=0,3-0,2=0,1\left(mol\right)\)
C% NaOH =\(\frac{0,1.40}{350}.100\%=1,14\%\)
Theo pthh
n\(_{Na2SO4}=n_{CuSO4}=0,1\left(mol\right)\)
C% CuSO4 =\(\frac{0,1.142}{350}.100\%=4,06\%\)
